A soap film on a ring of radius 3 cm evaporates. Light of wavelength 580 nm gives minimum transmission every 12 s. Re $fractive$ index is 1.45. The rate of evaporation is:
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A soap film on a ring of radius 3 cm evaporates. Light of wavelength 580 nm gives minimum transmission every 12 s. Re \fractive \fractive \fractive index is 1.45. The rate of evaporation is:

G1 Physics · Accepted Answer

Successive minima correspond to thickness change Δl=λ/(2μ). Thus |Δl/Δt|=580 nm/(2×1.45×12)=5/3 nm/s. Volume loss rate=πR²|Δl/Δt|=π(3×10⁻²)²(5/3×10⁻⁹)=15π×10⁻¹³ m³/s.